Answer: High pass filters, and in particular LC high pass filters are used in many RF applications where they block the lower frequencies and allow through higher frequency signals. Typically LC filters are used for the higher radio frequencies where active filters are not so manageable, and inductors more appropriate.
A) 350 J
- The initial internal energy of the cup is
- The final internal energy of the cup is
According to the first law of thermodynamics:
where
Q is the heat absorbed by the system
W is the work done on the system
The work done on the system in this case is 0, so we can rewrite the equation as
And so we find the heat transferred
B) IN the cup
Explanation:
in this situation, we see that the internal energy of the cup increases. The internal energy of an object/substance is proportional to its temperature, so it is a measure of the average kinetic energy of the molecules of the object/substance. Therefore, in this case, the temperature (and the energy of the molecules of the substance) has increased: this means that heat has been transferred INTO the system from the environment (the heat came from the sun).
The correct option is D.
Lumen is used to quantify the amount of total light energy that a source is putting out in all direction, thus, it refers to luminous output of a light source. Initial lumen refers to the luminosity of a light when it was first turned on; the luminosity is highest at this point.
Answer:
229,098.96 J
Explanation:
mass of water (m) = 456 g = 0.456 kg
initial temperature (T) = 25 degrees
final temperature (t) = - 10 degrees
specific heat of ice = 2090 J/kg
latent heat of fusion =33.5 x 10^(4) J/kg
specific heat of water = 4186 J/kg
for the water to be converted to ice it must undergo three stages:
- the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp
Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J
- the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp
Q = 0.456 x 33.5 x 10^(4) = 152760 J
- the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp
Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J
The quantity of heat removed from all three stages would be added to get the total heat removed.
Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J
The answer for this would be B!!
